# Is the convex hull of a compact set compact?

Let $$V$$ be a normed vector space and $$K$$ a compact subset of $$V$$. Is the convex hull of $$K$$ given by $$\langle K \rangle = \left\{\sum_{i=1}^n t_i x_i\mid x_i\in K, t_i≥0\text{ s.t. }\sum_i t_i=1\right\}$$ again compact?

• Possibly related: mathoverflow.net/questions/156321/convex-hulls-of-compact-sets. – Martin R Nov 16 '16 at 17:47
• @MartinR thanks, from the link in that question, the answer is no. – s.harp Nov 16 '16 at 17:54
• I don't think your definition of convex hull is correct. If $K=\{x,y,z\}$, is $\frac{x+y+z}{3} \in \langle K \rangle$? – Thomas Aug 25 '17 at 22:38
• @Thomas you are right. We obviously want want $\sum_i t_i\,x_i$ where $\sum_i t_i =1, t_i≥0$ or with what I have written for $\langle\cdot\rangle$: $$\bigcup_n \underbrace{\langle ... \langle}_{n} K\underbrace{\rangle...\rangle}_{n}$$Alternatively the smallest set containing $K$ so that $\langle X\rangle =X$ with the definition of $\langle \cdot\rangle$ that I have written. – s.harp Aug 27 '17 at 12:09

Consider $$u_n=(\underbrace{0,...,0}_{n-1},1/n,0,...)$$ and $$K=\bigcup_n \{u_n\} \cup \{0\}$$ a compact subset of $$\mathscr l^p(\mathbb N)$$. The convex hull of $$K$$ is given by elements of the form: $$\sum_{n=1}^k a_n u_{n}\qquad\text{s.t.:}\quad \sum_{n=1}^k a_n≤1\qquad a_n≥0$$ So also $$\sum_{n=1}^k 2^{-n}u_n$$ lies in it. But this sequence converges to $$\sum_{n=1}^\infty 2^{-n}u_n$$ which does not lie in it.